In the previous example, how many ml of 0.1M KMN04 are required to react with 30ml of 0.05M H202? The answer 3ml
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To solve this problem, you'll need to use the concept of stoichiometry and the balanced chemical equation for the reaction between KMnO4 and H2O2. Here is how you can calculate the answer:
Step 1: Write the balanced chemical equation for the reaction between KMnO4 and H2O2:
2 KMnO4 + 3 H2O2 → 2 MnO2 + 3 O2 + 2 KOH + 2 H2O
Step 2: Determine the stoichiometry of the reaction. From the balanced equation, we can see that 2 moles of KMnO4 react with 3 moles of H2O2.
Step 3: Calculate the number of moles of H2O2 in the given solution:
Moles of H2O2 = (Volume of H2O2 solution in liters) x (concentration of H2O2 in moles per liter)
Moles of H2O2 = 0.030 L x 0.05 mol/L = 0.0015 mol H2O2
Step 4: Use stoichiometry to determine the number of moles of KMnO4 required to react with the given number of moles of H2O2:
Moles of KMnO4 = (Moles of H2O2) x (Moles of KMnO4 / Moles of H2O2)
Moles of KMnO4 = 0.0015 mol x (2 mol KMnO4 / 3 mol H2O2) = 0.001 mol KMnO4
Step 5: Convert the moles of KMnO4 to milliliters using its molarity:
Volume of KMnO4 = (Moles of KMnO4) / (concentration of KMnO4)
Volume of KMnO4 = 0.001 mol / 0.1 mol/L = 0.01 L = 10 mL
Therefore, the answer is 10 mL, not 3 mL.